If the coefficient of kinetic friction is 0.22, how much work do you do when you slide a 50 kg box at constant

If the coefficient of kinetic friction is 0.22, how much work do you do when you slide a 50 kg box at constant speed across a 5.8m wide room?

(50 x 9.8) = weight of 490N.
Friction = (490 x 0.22) = 107.8N. force needed to move the box at a steady speed.
(107.8 x 5.8) = 625.24 Joules work done.

Well work equals the force times the displacment the displacement is 5.8m so that's easy. And the force is the friction multiplied by the normal force, which is the force of gravity or 50kg x 9.80m/s2 anyways do all that crap out and see what you get because I'm not doing it